Verified Solution • 2025–26 Session moderate

QJEEAdv15Ph91b0: JEE Advanced 2015 Physics - Class 12 JEEAdv15Ph91b0 Solution: For a radioactive material, its activity A and rate of change…

For a radioactive material, its activity A and rate of change of its activity R are defined as A=−dNdtA = - {{dN} \over {dt}} and R=−dAdtR = - {{dA} \over {dt}} , where N(t) is the number of nuclei at time t. Two radioactive source P(mean life τ\tau ) and Q (mean life 2 τ\tau ) have the same activity at t = 0. Their rate of change of activities at t = 2 τ\tau are RP and RQ, respectively. If RPRQ=ne{{{R_P}} \over {{R_Q}}} = {n \over e} , then the value of n is

Step-by-Step Verified Solution

Law of radioactivity : N=N0e−λtN = {N_0}{e^{ - \lambda t}} where λ\lambda = decay constant

Activity ∣A∣=∣−dNdt∣=N0λe−λt\left| A \right| = \left| {{{ - dN} \over {dt}}} \right| = {N_0}\lambda {e^{ - \lambda t}}

Rate of activity R=d∣A∣dt=N0λ2e−λtR = {{d|A|} \over {dt}} = {N_0}{\lambda ^2}{e^{ - \lambda t}}

At t = 0, A1 = A2. Therefore,

NOPλP=NOQλQ{N_{OP}}{\lambda _P} = {N_{OQ}}{\lambda _Q}

At t=2τ, RPRQ=(λPλQ)2(NOPNOQ)e−λP(25)e−λQ(25)=λPλQe(λQ−λP)25t = 2\tau ,\,{{{R_P}} \over {{R_Q}}} = {\left( {{{{\lambda _P}} \over {{\lambda _Q}}}} \right)^2}\left( {{{{N_{OP}}} \over {{N_{OQ}}}}} \right){{{e^{ - \lambda P(25)}}} \over {{e^{ - \lambda Q(25)}}}} = {{{\lambda _P}} \over {{\lambda _Q}}}{e^{({\lambda _Q} - {\lambda _P})25}}

Since mean life is given by τ=1λ\tau = {1 \over \lambda } .

Therefore,

RPRQ=λPλQe(125−15)25=λPλQe−1{{{R_P}} \over {{R_Q}}} = {{{\lambda _P}} \over {{\lambda _Q}}}{e^{{{\left( {{1 \over {25}} - {1 \over 5}} \right)}^{25}}}} = {{{\lambda _P}} \over {{\lambda _Q}}}{e^{ - 1}}

RPRQ=λ{λQ1e=ne{{{R_P}} \over {{R_Q}}} = {{{\lambda _\{ }} \over {{\lambda _Q}}}{1 \over e} = {n \over e}

n=λPλQ=2ττ=2n = {{{\lambda _P}} \over {{\lambda _Q}}} = {{2\tau } \over \tau } = 2

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